# How do you find all the real and complex roots of x^7 + 128 = 0?

Jan 12, 2016

${x}^{7} + 128 = {\prod}_{k = 0}^{6} \left(x + 2 \cos \left(\frac{2 k \pi}{7}\right) + 2 i \sin \left(\frac{2 k \pi}{7}\right)\right)$

Hence roots:

$x = - 2 \cos \left(\frac{2 k \pi}{7}\right) - 2 i \sin \left(\frac{2 k \pi}{7}\right)$

where $k = 0 , 1 , 2 , 3 , 4 , 5 , 6$

#### Explanation:

${x}^{7} + 128 = {x}^{7} + {2}^{7}$

By De Moivre's formula we find:

${\left(\cos \left(\frac{2 k \pi}{7}\right) + i \sin \left(\frac{2 k \pi}{7}\right)\right)}^{7} = \cos \left(2 k \pi\right) + i \sin \left(2 k \pi\right) = 1$

for any integer $k$.

So the $7$th roots of $1$ can be written as:

$\cos \left(\frac{2 k \pi}{7}\right) + i \sin \left(\frac{2 k \pi}{7}\right)$ where $k = 0 , 1 , 2 , 3 , 4 , 5 , 6$

Hence we find:

${x}^{7} + {2}^{7} = {\prod}_{k = 0}^{6} \left(x + 2 \cos \left(\frac{2 k \pi}{7}\right) + 2 i \sin \left(\frac{2 k \pi}{7}\right)\right)$

Jan 13, 2016

$z = 2 c i s \left(\pi + \frac{2 \pi k}{7}\right)$ for $k = 0 , 1 , 2 , 3 , 4 , 5 , 6$

#### Explanation:

We have

${x}^{7} + 128 = 0$

Numerically we know that $- 2$ is one of the roots.
$z = - 2$ has modulo $2$ and $\theta = \pi$, so we have the trigonometric representation

$z = 2 c i s \left(\pi\right)$

(Where $c i s \left(\theta\right) = \cos \left(\theta\right) + i \sin \left(\theta\right)$)

But the thing is, whenever we take the nth root of a number, we bissect the circle of $\sqrt[n]{| z |}$ radius in $n$ parts of equal distance. I.e.: if we take the square roots, the two roots will be exactly $\frac{2 \pi}{2}$ degrees apart, if we take the cubic root, they'll be $\frac{2 \pi}{3}$ apart and so on.

That means that we can derive that the general form of the roots are

$z = 2 c i s \left(\pi + \frac{2 \pi k}{7}\right)$ for $k = 0 , 1 , 2 , 3 , 4 , 5 , 6$ as when we have $k = 7$, we go back to the same root we already had.

Which we can easily check back using De Moivre's formula, since

${z}^{7} = 128 c i s \left(7 \pi + 7 \frac{2 \pi k}{7}\right) = 128 c i s \left(7 \pi + 2 \pi k\right)$

For any of the mentioned $k$, adding or subtracting $2 \pi$ is the same as doing nothing so

${z}^{7} = 128 c i s \left(7 \pi\right)$

And since $7 \pi = 6 \pi + \pi$

${z}^{7} = 128 c i s \left(\pi\right) = - 128$